Homework - Chapter 9

Conceptual

1. This problem involves hyperplanes in two dimensions.

1. Sketch the hyperplane 1 + 3X1 − X2 = 0. Indicate the set of points for which 1+3X1 −X2 > 0, as well as the set of points for which 1 + 3X1 − X2 < 0.
2. On the same plot, sketch the hyperplane −2 + X1 + 2X2 = 0. Indicate the set of points for which −2 + X1 + 2X2 > 0, as well as the set of points for which −2 + X1 + 2X2 < 0.

x1 = seq(-10,10)
x2_1 = 1 + 3*x1
x2_2 = 1 - ((1/2)*x1)
matplot(x = x1, y=cbind(x2_1,x2_2),type="l",ylab="X2")
text(x=-3,y=20,labels="(a) 1+3X1 −X2 < 0")
text(x=0,y=-20,labels="(a) 1+3X1 −X2 > 0")
text(x=-7,y=-9,labels="(b) −2 + X1 + 2X2 < 0",col="red")
text(x=6,y=5,labels="(b) −2 + X1 + 2X2 > 0",col="red")

2. We have seen that in p = 2 dimensions, a linear decision boundary takes the form β0 +β1X1 +β2X2 = 0. We now investigate a non-linear decision boundary.

1. Sketch the curve (1+X1)^2 +(2−X2)^2 =4.
2. On your sketch, indicate the set of points for which (1+X1)2 +(2−X2)2 >4, as well as the set of points for which (1+X1)2 +(2−X2)2 ≤4.

plot(NA,NA,type="n",xlim=c(-5,3),ylim=c(-1,5),xlab="X1",ylab="X2",asp=1)
symbols(c(-1),c(2),circles=c(2),add=T,inches=F)

predF = function(x1,x2){
   (1+x1)^2 + (2-x2)^2 - 4
}

predF(-1,2)

## [1] -4

predF(-4,2)

## [1] 5

predF(-1,5)

## [1] 5

text(x=-1,y=2,labels="< 4")
text(x=-4,y=2,labels="> 4")

3. Suppose that a classifier assigns an observation to the blue class if (1+X1)2 +(2−X2)2 >4, and to the red class otherwise. To what class is the observation (0, 0) classified? (−1, 1)? (2, 2)? (3, 8)?

ifelse(predF(0,0)>0,"Blue","Red")

## [1] "Blue"

ifelse(predF(-1,1)>0,"Blue","Red")

## [1] "Red"

ifelse(predF(2,2)>0,"Blue","Red")

## [1] "Blue"

ifelse(predF(3,8)>0,"Blue","Red")

## [1] "Blue"

4. Argue that while the decision boundary in (c) is not linear in terms of X1 and X2, it is linear in terms of X1, X12, X2, and X2.

The decision boundary can be rewritten as a function of the parameters, beta, that just treat X1^2 and X2^2 as features. So, if we pre-compute/evaluate those features ahead of time, then they aren’t part of the function, and the function that’s being fit is a linear additive function.

3. Here we explore the maximal margin classifier on a toy data set.

1. We are given n = 7 observations in p = 2 dimensions. For each observation, there is an associated class label.
2. Sketch the optimal separating hyperplane, and provide the equation for this hyperplane (of the form (9.1)).
3. Describe the classification rule for the maximal margin classifier. It should be something along the lines of “Classify to Red if β0 + β1X1 + β2X2 > 0, and classify to Blue otherwise.” Provide the values for β0, β1, and β2. Classify to Red if -.5 + (1)x1 + (-1)x2 < 0, and Blue otherwise.
4. On your sketch, indicate the margin for the maximal margin hyperplane.
5. Indicate the support vectors for the maximal margin classifier.
6. Argue that a slight movement of the seventh observation would not affect the maximal margin hyperplane. If that point moves anwhere under the dashed lower margin line, the support vectors remain the same.
7. Sketch a hyperplane that is not the optimal separating hyper- plane, and provide the equation for this hyperplane.
8. Draw an additional observation on the plot so that the two classes are no longer separable by a hyperplane.

x1 = c(3,2,4,1,2,4,4)
x2 = c(4,2,4,4,1,3,1)
y = c("Red","Red","Red","Red","Blue","Blue","Blue")

plot(x1,x2,col=y)
abline(-.5,1)
text(x=3,y=2.25,labels="0=x1-x2-0.5")

abline(0,1,lty=2)
abline(-1,1,lty=2)

points(2,1,pch=4,col="blue",cex=2)
points(2,2,pch=4,col="red",cex=2)
points(4,3,pch=4,col="blue",cex=2)
points(4,4,pch=4,col="red",cex=2)

abline(-.5,1.1,col="green",lty=2)
text(x=3,y=3,labels="0=1.1x1-x2-0.5",col="green")

points(x=2,y=2.5,col="blue")

Applied

4. Generate a simulated two-class data set with 100 observations and two features in which there is a visible but non-linear separation be- tween the two classes. Show that in this setting, a support vector machine with a polynomial kernel (with degree greater than 1) or a radial kernel will outperform a support vector classifier on the train- ing data. Which technique performs best on the test data? Make plots and report training and test error rates in order to back up your assertions.

library(e1071)

set.seed(1)
x = matrix(rnorm(100*2),ncol=2)
x[1:50,]=x[1:50,]-1
x[51:100,]=x[51:100,]+.5
y = c(rep(1,50),rep(2,50))
plot(x,col=y)

dat = data.frame(x=x,y=as.factor(y))
train = sample(100,50)
dat.train = dat[train,]
dat.test = dat[-train,]

fitsvm.linear = svm(y~.,data=dat.train,kernel="linear",cost=10000)
plot(fitsvm.linear,dat.train)

table(pred=fitsvm.linear$fitted,y=dat.train$y)

##     y
## pred  1  2
##    1 21  1
##    2  1 27

mean(fitsvm.linear$fitted != dat.train$y)

## [1] 0.04

fitsvm.poly = svm(y~.,data=dat.train,kernel="polynomial",degree=3,cost=10000)
plot(fitsvm.poly,dat.train)

mean(fitsvm.poly$fitted != dat.train$y)

## [1] 0

fitsvm.rad = svm(y~.,data=dat.train,kernel="radial",gamma=1,cost=10000)
plot(fitsvm.rad,dat.train)

mean(fitsvm.rad$fitted != dat.train$y)

## [1] 0

pred.linear = predict(fitsvm.linear,dat.test)
pred.poly = predict(fitsvm.poly,dat.test)
pred.rad = predict(fitsvm.rad,dat.test)

mean(pred.linear != dat.test$y)

## [1] 0.18

mean(pred.poly != dat.test$y)

## [1] 0.24

mean(pred.rad != dat.test$y)

## [1] 0.28

5. We have seen that we can fit an SVM with a non-linear kernel in order to perform classification using a non-linear decision boundary. We will now see that we can also obtain a non-linear decision boundary by performing logistic regression using non-linear transformations of the features.

1. Generate a data set with n = 500 and p = 2, such that the obser- vations belong to two classes with a quadratic decision boundary between them. For instance, you can do this as follows:

set.seed(1)
x1=runif(500)-0.5
x2=runif(500)-0.5
y=1*(x1^2-x2^2 > 0)

2. Plot the observations, colored according to their class labels. Your plot should display X1 on the x-axis, and X2 on the y- axis.

plot(x1,x2,col=(1+y))

3. Fit a logistic regression model to the data, using X1 and X2 as predictors.
4. Apply this model to the training data in order to obtain a pre- dicted class label for each training observation. Plot the ob- servations, colored according to the predicted class labels. The decision boundary should be linear.

dat = data.frame(x1=x1,x2=x2,y=as.factor(y))
fit.glm = glm(y~.,data=dat,family="binomial")
summary(fit.glm)

## 
## Call:
## glm(formula = y ~ ., family = "binomial", data = dat)
## 
## Deviance Residuals: 
##    Min      1Q  Median      3Q     Max  
## -1.179  -1.139  -1.112   1.206   1.257  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.087260   0.089579  -0.974    0.330
## x1           0.196199   0.316864   0.619    0.536
## x2          -0.002854   0.305712  -0.009    0.993
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 692.18  on 499  degrees of freedom
## Residual deviance: 691.79  on 497  degrees of freedom
## AIC: 697.79
## 
## Number of Fisher Scoring iterations: 3

contrasts(dat$y)

##   1
## 0 0
## 1 1

fit.glm.pred = ifelse(fit.glm$fitted.values >.5,1,0)
plot(dat$x1,dat$x2,col=(fit.glm.pred+1))

5. Now fit a logistic regression model to the data using non-linear functions of X1 and X2 as predictors (e.g. X12, X1 ×X2, log(X2), and so forth).
6. Apply this model to the training data in order to obtain a pre- dicted class label for each training observation. Plot the ob- servations, colored according to the predicted class labels. The decision boundary should be obviously non-linear. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear.

fit.glm.nonl = glm(y~poly(x1,2)+poly(x2,2)+(x1*x2),data=dat,family="binomial")

## Warning: glm.fit: algorithm did not converge

## Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

summary(fit.glm.nonl)

## 
## Call:
## glm(formula = y ~ poly(x1, 2) + poly(x2, 2) + (x1 * x2), family = "binomial", 
##     data = dat)
## 
## Deviance Residuals: 
##        Min          1Q      Median          3Q         Max  
## -8.240e-04  -2.000e-08  -2.000e-08   2.000e-08   1.163e-03  
## 
## Coefficients: (2 not defined because of singularities)
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -102.2     4302.0  -0.024    0.981
## poly(x1, 2)1   2715.3   141109.5   0.019    0.985
## poly(x1, 2)2  27218.5   842987.2   0.032    0.974
## poly(x2, 2)1   -279.7    97160.4  -0.003    0.998
## poly(x2, 2)2 -28693.0   875451.3  -0.033    0.974
## x1                 NA         NA      NA       NA
## x2                 NA         NA      NA       NA
## x1:x2          -206.4    41802.8  -0.005    0.996
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 6.9218e+02  on 499  degrees of freedom
## Residual deviance: 3.5810e-06  on 494  degrees of freedom
## AIC: 12
## 
## Number of Fisher Scoring iterations: 25

fit.glm.nonl.pred = ifelse(fit.glm.nonl$fitted.values > .5,1,0)
plot(dat$x1,dat$x2,col=(fit.glm.nonl.pred+1))

7. Fit a support vector classifier to the data with X1 and X2 as predictors. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.fit.linear = svm(y~.,data=dat,kernel="linear",cost=10)
summary(svm.fit.linear)

## 
## Call:
## svm(formula = y ~ ., data = dat, kernel = "linear", cost = 10)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  10 
##       gamma:  0.5 
## 
## Number of Support Vectors:  487
## 
##  ( 239 248 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  0 1

plot(svm.fit.linear,dat)

svm.pred.linear = svm.fit.linear$fitted
plot(dat$x1,dat$x2,col=(svm.pred.linear))

8. Fit a SVM using a non-linear kernel to the data. Obtain a class prediction for each training observation. Plot the observations, colored according to the predicted class labels.

svm.fit.rad = svm(y~., data=dat, kernel="radial",cost=10,gamma=1)
svm.pred.rad = svm.fit.rad$fitted
plot(dat$x1,dat$x2,col=svm.pred.rad)

1. Comment on your results. Logistic regression and SVM with a radial kernel provided very good fits, but the linear SVM was terrible, just classigying everything to 0.

6. At the end of Section 9.6.1, it is claimed that in the case of data that is just barely linearly separable, a support vector classifier with a small value of cost that misclassifies a couple of training observations may perform better on test data than one with a huge value of cost that does not misclassify any training observations. You will now investigate this claim.

1. Generate two-class data with p = 2 in such a way that the classes are just barely linearly separable.

set.seed(1)
#x = matrix(c(rnorm(50,3),rnorm(50),rnorm(50,3.5),rnorm(50)),ncol=2)
x = matrix(runif(1000*2),ncol=2)
x[1:500,2]=x[1:500,2]-1
x[501:1000,]=x[501:1000,]
y = c(rep(1,500),rep(2,500))
plot(x,col=y)

dat = data.frame(x=x,y=as.factor(y))

cost=c(1e-1,1e0,1e1,1e2,1e3,1e4,1e5)
tune.out = tune(svm, y~., data=dat, kernel="linear",
                ranges=list(cost=cost))

summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##   cost
##  10000
## 
## - best performance: 0 
## 
## - Detailed performance results:
##    cost error  dispersion
## 1 1e-01 0.003 0.004830459
## 2 1e+00 0.003 0.004830459
## 3 1e+01 0.003 0.004830459
## 4 1e+02 0.002 0.004216370
## 5 1e+03 0.002 0.004216370
## 6 1e+04 0.000 0.000000000
## 7 1e+05 0.000 0.000000000

cvCostErrors = data.frame(cost=cost,error=tune.out$performances$error,misclass=tune.out$performances$error*1000,src="cv")

trainCostErrors = data.frame(cost=cost,error=rep(0,length(cost)),misclass=rep(0,length(cost)),src="train")
for(i in 1:length(cost)){
   c = cost[i]
   fitsvm = svm(y~.,data=dat,kernel="linear",cost=c)
   predsvm = fitsvm$fitted
   trainCostErrors[i,"misclass"] = sum(predsvm != dat$y)
   trainCostErrors[i,"error"] = mean(predsvm != dat$y)
}

3. Generate an appropriate test data set, and compute the test errors corresponding to each of the values of cost considered. Which value of cost leads to the fewest test errors, and how does this compare to the values of cost that yield the fewest training errors and the fewest cross-validation errors?
4. Discuss your results.

#x = matrix(c(rnorm(50,3),rnorm(50),rnorm(50,3.5),rnorm(50)),ncol=2)
xtest = matrix(runif(1000*2),ncol=2)
xtest[1:500,2]=xtest[1:500,2]-1
xtest[501:1000,]=xtest[501:1000,]
ytest = c(rep(1,500),rep(2,500))
plot(xtest,col=ytest)

datTest = data.frame(x=xtest,y=as.factor(ytest))

cost=c(1e-1,1e0,1e1,1e2,1e3,1e4,1e5)

testCostErrors = data.frame(cost=cost,error=rep(0,length(cost)),misclass=rep(0,length(cost)),src="test")
for(i in 1:length(cost)){
   c = cost[i]
   fitsvm = svm(y~.,data=dat,kernel="linear",cost=c)
   predsvm = predict(fitsvm,newdata=datTest)
   testCostErrors[i,"misclass"] = sum(predsvm != datTest$y)
   testCostErrors[i,"error"] = mean(predsvm != datTest$y)
}

costErrors = rbind(cvCostErrors,testCostErrors,trainCostErrors)

plot(x=cost,y=costErrors[costErrors$src=="train","misclass"],type="l",xlab="cost",ylab="misclass",ylim=c(0,10))
points(x=cost,y=costErrors[costErrors$src=="cv","misclass"],type="l",col=2)
points(x=cost,y=costErrors[costErrors$src=="test","misclass"],type="l",col=3)

In my situation, cross validation and training error seemed to agree with one another that you should increase the cost to around 10,000 to get the best fit with 0 training errors. However, the test set showed that the lower cost of 1 performed the best, with only 4 training errors.

7. In this problem, you will use support vector approaches in order to predict whether a given car gets high or low gas mileage based on the Auto data set.

1. Create a binary variable that takes on a 1 for cars with gas mileage above the median, and a 0 for cars with gas mileage below the median.

library(ISLR)
library(e1071)
data(Auto)

Auto$mpgHigh = as.factor(ifelse(Auto$mpg > median(Auto$mpg), 1, 0))

2. Fit a support vector classifier to the data with various values of cost, in order to predict whether a car gets high or low gas mileage. Report the cross-validation errors associated with different values of this parameter. Comment on your results

set.seed(42)
tune.out = tune(svm, mpgHigh~.-mpg, data=Auto, kernel="linear", 
               ranges=list(cost=c(1e-2,1e-1,1e0,1e1,1e2,1e3,1e4,1e5)))

summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##     1
## 
## - best performance: 0.08685897 
## 
## - Detailed performance results:
##    cost      error dispersion
## 1 1e-02 0.08923077 0.04999616
## 2 1e-01 0.09423077 0.06011835
## 3 1e+00 0.08685897 0.04239744
## 4 1e+01 0.09955128 0.04258955
## 5 1e+02 0.11743590 0.04404592
## 6 1e+03 0.10480769 0.05218686
## 7 1e+04 0.10730769 0.04678970
## 8 1e+05 0.10730769 0.04678970

It looks like the cost of 1 yielded the best performance using cross validation. (c) Now repeat (b), this time using SVMs with radial and polyno- mial basis kernels, with different values of gamma and degree and cost. Comment on your results.

tune.out.poly = tune(svm, mpgHigh~.-mpg, data=Auto, kernel="polynomial",
                     ranges=list(cost=c(1e-2,1e-1,1e0,1e1,1e2,1e3,1e4,1e5),
                                 degree=c(2,3,4,5)))
summary(tune.out.poly)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##   cost degree
##  1e+05      3
## 
## - best performance: 0.1021795 
## 
## - Detailed performance results:
##     cost degree     error dispersion
## 1  1e-02      2 0.5434615 0.03101869
## 2  1e-01      2 0.5434615 0.03101869
## 3  1e+00      2 0.5434615 0.03101869
## 4  1e+01      2 0.5436538 0.06825206
## 5  1e+02      2 0.3088462 0.05000621
## 6  1e+03      2 0.2781410 0.04619187
## 7  1e+04      2 0.1608333 0.06617894
## 8  1e+05      2 0.1914744 0.06973807
## 9  1e-02      3 0.5434615 0.03101869
## 10 1e-01      3 0.5434615 0.03101869
## 11 1e+00      3 0.5434615 0.03101869
## 12 1e+01      3 0.5434615 0.03101869
## 13 1e+02      3 0.3880128 0.12149162
## 14 1e+03      3 0.2604487 0.05726819
## 15 1e+04      3 0.1685256 0.05971483
## 16 1e+05      3 0.1021795 0.04036958
## 17 1e-02      4 0.5434615 0.03101869
## 18 1e-01      4 0.5434615 0.03101869
## 19 1e+00      4 0.5434615 0.03101869
## 20 1e+01      4 0.5434615 0.03101869
## 21 1e+02      4 0.5434615 0.03101869
## 22 1e+03      4 0.5434615 0.03101869
## 23 1e+04      4 0.4390385 0.07173917
## 24 1e+05      4 0.3370513 0.06165839
## 25 1e-02      5 0.5434615 0.03101869
## 26 1e-01      5 0.5434615 0.03101869
## 27 1e+00      5 0.5434615 0.03101869
## 28 1e+01      5 0.5434615 0.03101869
## 29 1e+02      5 0.5434615 0.03101869
## 30 1e+03      5 0.5434615 0.03101869
## 31 1e+04      5 0.5434615 0.03101869
## 32 1e+05      5 0.4313462 0.07316837

tune.out.rad = tune(svm, mpgHigh~.-mpg, data=Auto, kernel="radial",
                     ranges=list(cost=c(1e-2,1e-1,1e0,1e1,1e2,1e3,1e4,1e5),
                                 gamma=c(.1,.5,1,2,3,5)))
summary(tune.out.rad)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost gamma
##     1   0.5
## 
## - best performance: 0.08166667 
## 
## - Detailed performance results:
##     cost gamma      error dispersion
## 1  1e-02   0.1 0.22967949 0.09904027
## 2  1e-01   0.1 0.08679487 0.04021093
## 3  1e+00   0.1 0.08679487 0.04021093
## 4  1e+01   0.1 0.09448718 0.03648306
## 5  1e+02   0.1 0.13019231 0.05744045
## 6  1e+03   0.1 0.12256410 0.05116600
## 7  1e+04   0.1 0.12256410 0.05116600
## 8  1e+05   0.1 0.12256410 0.05116600
## 9  1e-02   0.5 0.54083333 0.03353278
## 10 1e-01   0.5 0.08935897 0.04032884
## 11 1e+00   0.5 0.08166667 0.04463634
## 12 1e+01   0.5 0.09455128 0.03438412
## 13 1e+02   0.5 0.09967949 0.03527551
## 14 1e+03   0.5 0.09967949 0.03527551
## 15 1e+04   0.5 0.09967949 0.03527551
## 16 1e+05   0.5 0.09967949 0.03527551
## 17 1e-02   1.0 0.54083333 0.03353278
## 18 1e-01   1.0 0.54083333 0.03353278
## 19 1e+00   1.0 0.08423077 0.03991005
## 20 1e+01   1.0 0.08685897 0.03667161
## 21 1e+02   1.0 0.08685897 0.03667161
## 22 1e+03   1.0 0.08685897 0.03667161
## 23 1e+04   1.0 0.08685897 0.03667161
## 24 1e+05   1.0 0.08685897 0.03667161
## 25 1e-02   2.0 0.54083333 0.03353278
## 26 1e-01   2.0 0.54083333 0.03353278
## 27 1e+00   2.0 0.12006410 0.03676959
## 28 1e+01   2.0 0.11743590 0.03272065
## 29 1e+02   2.0 0.11743590 0.03272065
## 30 1e+03   2.0 0.11743590 0.03272065
## 31 1e+04   2.0 0.11743590 0.03272065
## 32 1e+05   2.0 0.11743590 0.03272065
## 33 1e-02   3.0 0.54083333 0.03353278
## 34 1e-01   3.0 0.54083333 0.03353278
## 35 1e+00   3.0 0.36961538 0.13200605
## 36 1e+01   3.0 0.33634615 0.11091454
## 37 1e+02   3.0 0.33634615 0.11091454
## 38 1e+03   3.0 0.33634615 0.11091454
## 39 1e+04   3.0 0.33634615 0.11091454
## 40 1e+05   3.0 0.33634615 0.11091454
## 41 1e-02   5.0 0.54083333 0.03353278
## 42 1e-01   5.0 0.54083333 0.03353278
## 43 1e+00   5.0 0.47698718 0.04089684
## 44 1e+01   5.0 0.47698718 0.03513181
## 45 1e+02   5.0 0.47698718 0.03513181
## 46 1e+03   5.0 0.47698718 0.03513181
## 47 1e+04   5.0 0.47698718 0.03513181
## 48 1e+05   5.0 0.47698718 0.03513181

The linear (cost = 1, error .086) and radial (gamma = .5, cost = 1, error .081) outperformed the best polynomial kernel svm (degree = 3, cost = 1e5, error .10). I’d say radial is the best choice here, because the low cost and gamma suggest that the likelihood of overfitting the data is lower than the high costs it took to achieve good cross validation performance using the other two kernel methods. (d) Make some plots to back up your assertions in (b) and (c). Hint: In the lab, we used the plot() function for svm objects only in cases with p = 2. When p > 2, you can use the plot() function to create plots displaying pairs of variables at a time. Essentially, instead of typing > plot(svmfit , dat) where svmfit contains your fitted model and dat is a data frame containing your data, you can type > plot(svmfit , dat , x1∼x4) in order to plot just the first and fourth variables. However, you must replace x1 and x4 with the correct variable names. To find out more, type ?plot.svm.

svmfit.linear = svm(mpgHigh~.,data=Auto,kernel="linear",cost=1)

plotSvmPairs = function(fit, var, noPlotList, dataSet) {
   for(name in names(dataSet)[!(names(dataSet) %in% c(noPlotList, var))]){
      print(paste0("Plotting ",var,"~",name))
      plot(fit, dataSet, as.formula(paste0(var,"~",name)))
   }
}

noPlotList = c("name","mpgHigh")
plotSvmPairs(fit=svmfit.linear, var="mpg", noPlotList=noPlotList, Auto)

## [1] "Plotting mpg~cylinders"

## [1] "Plotting mpg~displacement"

## [1] "Plotting mpg~horsepower"

## [1] "Plotting mpg~weight"

## [1] "Plotting mpg~acceleration"

## [1] "Plotting mpg~year"

## [1] "Plotting mpg~origin"

svmfit.poly = svm(mpgHigh~.,data=Auto,kernel="polynomial",cost=1e5,degree=3)
plotSvmPairs(fit=svmfit.poly, var="mpg", noPlotList=noPlotList, Auto)

## [1] "Plotting mpg~cylinders"

## [1] "Plotting mpg~displacement"

## [1] "Plotting mpg~horsepower"

## [1] "Plotting mpg~weight"

## [1] "Plotting mpg~acceleration"

## [1] "Plotting mpg~year"

## [1] "Plotting mpg~origin"

svm.radial <- svm(mpgHigh ~ ., data = Auto, kernel = "radial", cost = 100, gamma = 0.01)
plotSvmPairs(fit=svm.radial, var="mpg", noPlotList=noPlotList, Auto)

## [1] "Plotting mpg~cylinders"

## [1] "Plotting mpg~displacement"

## [1] "Plotting mpg~horsepower"

## [1] "Plotting mpg~weight"

## [1] "Plotting mpg~acceleration"

## [1] "Plotting mpg~year"

## [1] "Plotting mpg~origin"

8. This problem involves the OJ data set which is part of the ISLR package.

1. Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.

library(ISLR)
library(e1071)
data(OJ)

set.seed(42)
train = sample(nrow(OJ),800)

OJ.train = OJ[train,]
OJ.test = OJ[-train,]

2. Fit a support vector classifier to the training data using cost=0.01, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics, and describe the results obtained.

svmfit = svm(Purchase~., data=OJ.train, kernel="linear", cost=0.01)
summary(svmfit)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "linear", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  linear 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  439
## 
##  ( 219 220 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

The fit to the training data used 423 support vectors with 211 in one class and 212 in the other. (c) What are the training and test error rates?

print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.16"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.19"

4. Use the tune() function to select an optimal cost. Consider val- ues in the range 0.01 to 10.

set.seed(43)
cost = 10^seq(-2,1,by=.25)
tune.out = tune(svm, Purchase~., data = OJ.train, kernel="linear", ranges=list(cost=cost))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##  cost
##   0.1
## 
## - best performance: 0.16875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.17125 0.06347845
## 2   0.01778279 0.17125 0.06347845
## 3   0.03162278 0.17250 0.06449591
## 4   0.05623413 0.17250 0.06422616
## 5   0.10000000 0.16875 0.05781015
## 6   0.17782794 0.17125 0.06456317
## 7   0.31622777 0.17500 0.05921946
## 8   0.56234133 0.17375 0.05964304
## 9   1.00000000 0.17625 0.05964304
## 10  1.77827941 0.17375 0.06022239
## 11  3.16227766 0.17375 0.05935124
## 12  5.62341325 0.17375 0.05787019
## 13 10.00000000 0.17500 0.05892557

Best cost was 0.01. (e) Compute the training and test error rates using this new value for cost.

svmfit = svm(Purchase~., data=OJ.train, kernel="linear", cost=0.1)
print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.16"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.17"

6. Repeat parts (b) through (e) using a support vector machine with a radial kernel. Use the default value for gamma.

svmfit = svm(Purchase~., data=OJ.train, kernel="radial", cost=0.01)
summary(svmfit)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "radial", 
##     cost = 0.01)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  radial 
##        cost:  0.01 
##       gamma:  0.05555556 
## 
## Number of Support Vectors:  638
## 
##  ( 318 320 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.4"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.37"

set.seed(43)
cost = 10^seq(-2,1,by=.25)
tune.out = tune(svm, Purchase~., data = OJ.train, kernel="radial", ranges=list(cost=cost))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##       cost
##  0.5623413
## 
## - best performance: 0.1775 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39750 0.08815233
## 2   0.01778279 0.39750 0.08815233
## 3   0.03162278 0.34625 0.11815274
## 4   0.05623413 0.20875 0.07169815
## 5   0.10000000 0.18375 0.04825065
## 6   0.17782794 0.19000 0.05027701
## 7   0.31622777 0.18250 0.04090979
## 8   0.56234133 0.17750 0.04281744
## 9   1.00000000 0.17750 0.04556741
## 10  1.77827941 0.18125 0.04259385
## 11  3.16227766 0.18500 0.04440971
## 12  5.62341325 0.18125 0.04340139
## 13 10.00000000 0.19875 0.05447030

svmfit = svm(Purchase~., data=OJ.train, kernel="radial", cost=0.5623413)
print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.15"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.15"

7. Repeat parts (b) through (e) using a support vector machine with a polynomial kernel. Set degree=2.

svmfit = svm(Purchase~., data=OJ.train, kernel="polynomial", cost=0.01, degree=2)
summary(svmfit)

## 
## Call:
## svm(formula = Purchase ~ ., data = OJ.train, kernel = "polynomial", 
##     cost = 0.01, degree = 2)
## 
## 
## Parameters:
##    SVM-Type:  C-classification 
##  SVM-Kernel:  polynomial 
##        cost:  0.01 
##      degree:  2 
##       gamma:  0.05555556 
##      coef.0:  0 
## 
## Number of Support Vectors:  642
## 
##  ( 318 324 )
## 
## 
## Number of Classes:  2 
## 
## Levels: 
##  CH MM

print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.4"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.37"

set.seed(43)
cost = 10^seq(-2,1,by=.25)
tune.out = tune(svm, Purchase~., data = OJ.train, kernel="polynomial", degree=2, ranges=list(cost=cost))
summary(tune.out)

## 
## Parameter tuning of 'svm':
## 
## - sampling method: 10-fold cross validation 
## 
## - best parameters:
##      cost
##  5.623413
## 
## - best performance: 0.1875 
## 
## - Detailed performance results:
##           cost   error dispersion
## 1   0.01000000 0.39750 0.08815233
## 2   0.01778279 0.38625 0.08527936
## 3   0.03162278 0.37375 0.08446112
## 4   0.05623413 0.34750 0.07812917
## 5   0.10000000 0.33750 0.07660323
## 6   0.17782794 0.25625 0.07270803
## 7   0.31622777 0.22250 0.05974483
## 8   0.56234133 0.22125 0.05834821
## 9   1.00000000 0.21125 0.05905800
## 10  1.77827941 0.20500 0.06043821
## 11  3.16227766 0.19250 0.04794383
## 12  5.62341325 0.18750 0.03908680
## 13 10.00000000 0.18875 0.04059026

svmfit = svm(Purchase~., data=OJ.train, kernel="polynomial", degree=2, cost=5.62341325)
print(paste0("Training error: ", round(mean(svmfit$fitted != OJ.train$Purchase),2)))

## [1] "Training error: 0.16"

print(paste0("Test error: ", round(mean(predict(svmfit,OJ.test) != OJ.test$Purchase),2)))

## [1] "Test error: 0.17"

Overall, the linear kernel with cost = .1, and the polynomial kernel with cost = 5.62341325 and degree 2 both underperform the radial kernel with default gamma and cost = 0.5623413. The latter two achieved 17% test error rates, while the radial kernel was able to achieve 15%.